关于同余方程$x^2+xy+y^2 \equiv 0\pmod{n^2}$解特性的证明问询

Alright, let's break down how to prove this result step by step. The core strategy is to leverage the Chinese Remainder Theorem (CRT) and first analyze the equation modulo prime powers—since any integer ( n ) can be factored into prime powers, CRT lets us handle each prime power component independently. If we can show that for every prime power ( p^e ) (where ( p ) is a prime not congruent to 1 mod 6), the only solutions to ( x^2 + xy + y^2 \equiv 0 \pmod{p^{2e}} ) are ( x, y \equiv 0 \pmod{p^e} ), then combining these results via CRT will give us the conclusion for ( n^2 ).

Step 1: Analyze Each Prime Power Case #

Let's go through each allowed prime type (2, 3, and primes ( p \equiv 2 \pmod{3} )) one by one:

Case 1: Prime ( p = 2 ) #

First, check the base case modulo 4:

• Enumerate all possible ( x, y \mod 4 ):
  • If both are odd: ( 1 + 1 + 1 = 3 \not\equiv 0 \pmod{4} )
  • If one is odd, one even: ( 1 + 0 + 0 = 1 \not\equiv 0 \pmod{4} ) (or vice versa)
  • If both are even: ( 0 + 0 + 0 = 0 \pmod{4} )
    So the only solutions modulo 4 are when ( x, y ) are even.

Now use induction for higher powers of 2:

• Assume for some ( k \geq 1 ), all solutions to ( x^2 + xy + y^2 \equiv 0 \pmod{2^{2k}} ) satisfy ( x, y \equiv 0 \pmod{2^k} ). Let ( x = 2^k a ), ( y = 2^k b ). Substitute into the equation:
  [
  2^{2k}(a2 + ab + b^2) \equiv 0 \pmod{2^{2k+2}}
  ]
  Divide both sides by ( 2^{2k} ): ( a^2 + ab + b^2 \equiv 0 \pmod{4} ). From our base case, ( a, b ) must be even, so ( a = 2a' ), ( b = 2b' ). This gives ( x = 2^{k+1}a' ), ( y = 2^{k+1}b' ), meaning ( x, y \equiv 0 \pmod{2^{k+1}} ) for the modulo ( 2^{2(k+1)} ) case. The induction holds.

Case 2: Prime ( p = 3 ) #

Start with the base case modulo 9:

• Enumerate ( x, y \mod 3 ):
  • If either ( x ) or ( y ) is not divisible by 3:
    • If one is 0 mod 3, the other isn't: ( 0 + 0 + y^2 \equiv y^2 \not\equiv 0 \pmod{9} ) (or vice versa)
    • If both are 1 or 2 mod 3: ( 1+1+1=3 ), ( 4+2+4=10\equiv1 ), (1+2+4=7), (4+1+1=6)—none are 0 mod9
  • Only when both ( x, y \equiv 0 \pmod{3} ) do we get a solution modulo9.

Induction for higher powers of 3:

• Assume for ( k \geq1 ), solutions to ( x^2 + xy + y^2 \equiv0 \pmod{3^{2k}} ) satisfy ( x,y\equiv0\pmod{3^k} ). Let (x=3^k a), (y=3^k b). Substitute:
  [
  3^{2k}(a2+ab+b^{2)\equiv0\pmod{3}{2k+2}}
  ]
  Divide by (3^{2k}): (a^2+ab+b2\equiv0\pmod{9}). From the base case, (a,b\equiv0\pmod{3}), so (a=3a'), (b=3b'), giving (x=3^{k+1}a'), (y=3^{k+1}b'). The induction holds.

Case3: Primes ( p \equiv2\pmod{3} ) (odd primes not equal to3) #

First, handle the base case modulo ( p ):

• Suppose (x^2+xy+y2\equiv0\pmod{p}). If (x\equiv0\pmod{p}), then (y^2\equiv0\pmod{p}), so (y\equiv0\pmod{p}) (and vice versa). If (x,y\not\equiv0\pmod{p}), divide both sides by (y^2) (since (y) is invertible mod (p)) and let (t = x/y \pmod{p}). The equation becomes:
  [
  t^2 + t + 1 \equiv0\pmod{p}
  ]
  The discriminant of this quadratic is (1-4=-3). For this equation to have a solution, (-3) must be a quadratic residue modulo (p). Using quadratic reciprocity:
  • (\left(\frac{-3}{p}\right) = \left(\frac{-1}{p}\right)\left(\frac{3}{p}\right))
  • (\left(\frac{-1}{p}\right)=(-1)^{(p-1)/2}), and (\left(\frac{3}{p}\right)=\left(\frac{p}{3}\right)(-1)^{{(3-1)(p-1)/4}=\left(\frac{p}{3}\right)(-1)}{(p-1)/2})
  • Multiplying these gives (\left(\frac{-3}{p}\right)=\left(\frac{p}{3}\right)). Since (p\equiv2\pmod{3}), (\left(\frac{p}{3}\right)=\left(\frac{2}{3}\right)=-1), so (-3) is not a quadratic residue mod (p). Thus the quadratic has no solutions, meaning the only solutions modulo (p) are (x,y\equiv0\pmod{p}).

Now lift this result to (p^{2e}) using induction:

• Assume for (k\geq1), solutions to (x^2+xy+y2\equiv0\pmod{p^{2k}}) satisfy (x,y\equiv0\pmod{p^k}). Let (x=p^k a), (y=p^k b). Substitute:
  [
  p^{2k}(a2+ab+b^{2)\equiv0\pmod{p}{2k+2}}
  ]
  Divide by (p^{2k}): (a^2+ab+b2\equiv0\pmod{p^2}). From the base case for (p^2) (we can show similarly that only (a,b\equiv0\pmod{p}) work here), we get (a=pa'), (b=pb'), so (x=p^{k+1}a'), (y=p^{k+1}b'). The induction holds.

Step2: Combine Results via Chinese Remainder Theorem #

Since (n) factors into prime powers (n=\prod p_i^{e_i}), where each (p_i) is 2, 3, or ≡2 mod3, the equation (x^2+xy+y2\equiv0\pmod{n^2}) is equivalent to the system of equations modulo each (p_i^{2e_i}). From Step1, each system requires (x,y\equiv0\pmod{p_i^{e_i}}). By CRT, this implies (x,y\equiv0\pmod{n}), which is exactly what we needed to prove.

内容的提问来源于stack exchange，提问作者Gareth Ma